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Question Number 82582 by M±th+et£s last updated on 22/Feb/20

$$solve$$$$x^3-3x+1=0$$

Commented by mr W last updated on 23/Feb/20

$$x^3+3(-1)x+2\left(\frac{1}{2}\right)=0$$$$p=-1$$$$q=\frac{1}{2}$$$$\mathrm{\Delta }=p^3+q^2=-1+\frac{1}{4}=-\frac{3}{4}<0$$$$\Rightarrow threerealroots$$$$x=2\sqrt{-p}\sin (\frac{1}{3}{\sin }^{-1}\frac{q}{\sqrt{-p^3}}+\frac{2k\pi }{3})$$$$=2\sin (\frac{1}{3}{\sin }^{-1}\frac{1}{2}+\frac{2k\pi }{3})$$$$=2\sin (\frac{\pi }{18}+\frac{2k\pi }{3})(k=0,1,2)$$$$x_1=2\sin \left(\frac{\pi }{18}\right)=0.347296$$$$x_2=2\sin (\frac{\pi }{18}+\frac{2\pi }{3})=1.532089$$$$x_3=2\sin (\frac{\pi }{18}+\frac{4\pi }{3})=-1.879385$$

Commented by M±th+et£s last updated on 23/Feb/20

$$thankyousirnicesolution$$

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